```
This is a cross-post of someone else's question.
```

I am cross-posting this question from MSE since it hasn't received any answers.

In the paper Natural Operations on Differential Forms, the author R. Palais shows that the exterior derivative $d$ is characterized as the unique "natural" linear map from $\Phi^p$ to $\Phi^{p+1}$ (Palais' $\Phi^p$ is what is perhaps more commonly written as $\Omega^p$, and "commutes with all diffeomorphisms", I believe, means $f^*(d\omega) = d(f^*\omega)$):

the exterior derivative on $p$-forms is determined to within a scalar factor by the condition that it be a linear mapping into $p+1$ forms which commutes with all diffeomorphisms.

I've tried to read the proof in the paper, but I'm struggling to follow the details and missing a sense of the big picture of the proof.

I'm interested in Palais' claim because this characterization is the most compelling one I have seen $-$ it seems far more appropriate as an axiomatic definition of $d$ than the definitions found in many textbooks, which often define $d$ based on properties such as
$d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$ (where $\alpha$ is a $k$-form),
$d^2=0$,
or $d(\sum \omega_{i_1...i_k}dx_{i_1} \wedge \cdots \wedge dx_{i_k}) = \sum d\omega_{i_1...i_k} \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}$.
While these are indeed quite basic properties of $d$, they are more appropriate as theorems than as *a priori* assumptions. (Of course, what is more "natural" is a matter of opinion, so please don't belabor the issue.)

Since it's such an innocent-looking and natural characterization, I would like to see a clear, motivated, reasonably elementary proof of it. Why ought it to be true that there is only one natural linear map from $\Omega^p$ to $\Omega^{p+1}$, up to a constant multiple? What are the *key* steps of a proof?

thinkthe author means smooth maps. $\endgroup$ – Exterior Jan 10 '15 at 23:20